Keep rolling

Let $x_i$ , $1 \leq i \leq 6$ be the number of times $i$ appears after that the single student rolled the $9$ dices. We have that

$\displaystyle \sum_{i=1}^{6} x_i=9, \ x_i \geq 0$ .

In other terms, the single outcome $i$ can appear $0, \ 1,...$ or $9$ times out of $9$ roll. Despite the possible outcomes, the sum of the frequences has to be $9$ , since the student rolled exactly $9$ dices. The previous equation has $\displaystyle \binom{9+6-1}{6-1}=\binom{14}{5}=2002$ possible solution, applying a notorious stars and bars argument. Hence, even if the first $2002$ students will record $2002$ different solution, the minimum number of students who must have recorded the same result as another student is $2014-2002=\boxed{12}$ .